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Is there a way of approximating e^x for large xby Chain
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#1
Dec2212, 09:38 AM

P: 30

I'm trying to evaluate an integral with e^x where x is huge in the domain of the integral so I can't evaluate it numerically without making an approximation.



#2
Dec2212, 10:01 AM

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P: 11,917

Can you add more context? There might be a clever way to approximate e^(x), but if that value is not added to something, e^(x)=e^(xx_{0})*e^(x_{0}) where the second factor is independent of x and the first factor can be chosen to be about e^0.



#3
Dec2212, 10:15 AM

P: 30

The integral is 4*(r^2)*exp(2*r/a)/a^3 integrated between 1 and ∞ (The probability an electron in the ground state of hydrogen is more than 1 metre away from the nucleus) a=0.529*10^10



#4
Dec2212, 11:14 AM

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P: 16,091

Is there a way of approximating e^x for large x
Note that your integrand is one you can antidifferentiate without much trouble, so you can get the exact answer. (e.g. integration by parts. Or computer algebra package)



#5
Dec2212, 12:17 PM

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P: 11,917

If you just want an estimate: The exponential will drop very quick, so regions with r>1m+eps are irrelevant and r^2 is nearly constant and =1m^2. Therefore, the integral is simply 4m^2/a^3 * exp(2r/a) which can be evaluated as 2m^2/a^2 * exp(2m/a) ≈ 8*10^20 * exp(4*10^10) ≈ 10^(10^10) where the last approximation is very rough.



#6
Dec2212, 02:39 PM

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P: 6,076

e^{x} = 10^{0.43429448190325x}. The constant is log_{10}e.
Using the above for large x, you can separate the integer and fractional parts of the exponent. I assume you know how to proceed from here. 


#7
Dec2212, 09:46 PM

HW Helper
P: 2,264

Are you talking about Gauss–Laguerre quadrature?
[tex]\int_0^\infty f(x) e^{x} \mathop{\text{dx}}\sim \sum_{i=1}^n w_i f(x_i)[/tex] where xi are zeros of a Laguerre polynomial and [tex]w_i=\frac{x_i}{(n+1)^2[L_{n+1}(x_i)]^2}[/tex] 


#8
Dec2312, 02:23 AM

P: 30

Fair enough, yeah I realised after posting this the integral could be solved analytically >__< and I got a valule of something like 10^(10^10) but thanks for the responses :)



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